00:01
Here the problem says a stone is thrown vertically upwards from the base a of a 27 .60 meters clip.
00:12
If the distance by which ball clears the top of the clip is 8 .60 meters, determine the impact velocity at the top of the cliff.
00:26
Determine the total time for the ball land at the top of the cliff.
00:31
The clip so now here we are given with height of clip right we are given ball clear top of clip by 8 .60 meters maximum height covered by ball we will just calculate right so here this will be the height covered by the ball that is represented as let us say so this is 27 .60 plus 8 .60 yes because ball clears top of the clip so that is from the above of the clip right so let us say this is let us say this is clip right so this is 27 let us say so it is clearing 8 .6 meters 6 meter right so this is from the top of the clip right so that's why we are telling okay, so this is how that's why we have added this value so this is the height for the height covered from the ball by the ball.
01:56
So now let us assume that initial velocity of this tone is v1 let us say.
02:02
Okay.
02:03
So now the time to reach the edge, that is height covered by the ball.
02:11
Okay.
02:12
So from equation of motion we know.
02:15
So finally we need to find the value total time for the ball right we need to find this okay so from equation of motion we know that s is equal to u t plus half a t square right so here what we will do we will just put all the values like we have assumed initial velocity u is initial velocity instead of that we will put v1 as we have assumed here initial velocity of stone so it will be v1 t1 we have assumed initial velocity as initial time as t1 sorry not initial time this is time only so this is t1 so this will be plus half into 9 .81 into t1 square so i'm really sorry here we will put here this is opposite to the gravitational acceleration right so it will be minus only so 9 .81 meter per second square is the value of g and this will be t1 square so we will finally get it as s is equals to b1 t1 minus 4 .905 t1 square so this is our equation number 1 also we have one one one equation of motion is this and we can also find out from the another equation of motion as well so from again from the another equation of motion we know b square is equal to u squared plus 2 as so so here the final velocity is because it is coming to the ground and we have assumed v1 square as v as v1 as initial velocity so we even is square plus so this is again minus right because we are moving towards the against the gravitational so it is minus again 2 point 2 into 9 .81 meter per second square gravitationalization into height that is h so now it will give us v1 square equals to so here we only need to to put all the value so it will be 2 into 9 .81 and h is given by 36 .20 meter so this will finally give us v1 as 26.
06:30
6 by meter per second.
06:35
So this is our initial velocity we have got.
06:41
Now, we have assumed here equation 1.
06:45
So we will just put all the values, putting values in equation 1...